Problem: For a certain square, two vertices lie on the line $y = 2x - 17,$ and the other two vertices lie on the parabola $y = x^2.$  Find the smallest possible area of the square.
Answer: The two vertices that lie on $y = x^2$ must lie on a line of the form $y = 2x + k.$  Setting $y = x^2,$ we get $x^2 = 2x + k,$ so $x^2 - 2x - k = 0.$  Let $x_1$ and $x_2$ be the roots of this quadratic, so by Vieta's formulas, $x_1 + x_2 = 2$ and $x_1 x_2 = -k.$

The two vertices on the parabola are then $(x_1, 2x_1 + k)$ and $(x_2, 2x_2 + k),$ and the square of the distance between them is
\begin{align*}
(x_1 - x_2)^2 + (2x_1 - 2x_2)^2 &= 5(x_1 - x_2)^2 \\
&= 5[(x_1 + x_2)^2 - 4x_1 x_2] \\
&= 5 (4 + 4k) \\
&= 20(k + 1).
\end{align*}[asy]
unitsize(0.3 cm);

real parab (real x) {
  return(x^2);
}

pair A, B, C, D;

A = (-1,1);
B = (3,9);
C = (11,5);
D = (7,-3);

draw(graph(parab,-3.5,3.5));
draw(interp(D,C,-0.4)--interp(D,C,1.4));
draw(interp(A,B,-0.4)--interp(A,B,1.4));
draw(A--D);
draw(B--C);

label("$y = x^2$", (3.5,3.5^2), N);
label("$y = 2x - 17$", interp(D,C,1.4), N);
[/asy]

The point $(0,k)$ lies on the line $y = 2x + k,$ and its distance to the line $y - 2x + 17 = 0$ is
\[\frac{|k + 17|}{\sqrt{5}}.\]Hence,
\[20 (k + 1) = \frac{(k + 17)^2}{5}.\]This simplifies to $k^2 - 66k + 189 = 0,$ which factors as $(k - 3)(k - 63) = 0.$  Hence, $k = 3$ or $k = 63.$

We want to find the smallest possible area of the square, so we take $k = 3.$  This gives us $20(k + 1) = \boxed{80}.$